Lastly, we derive the degree of the polynomial \(Q_{N}^{[m]}(z)\)  from its definition (REF ). Notice from Eq. (REF ) that the highest-degree term of \(p^{[m]}_{k}(z)\)  is \(z^k/k!\) . Retaining only this highest-degree term of \(p^{[m]}_{k}(z)\)  in the determinant (REF ) for \(Q_{N}^{[m]}(z)\)  and evaluating the simplified determinant by the same technique as that used in Ref. [1]}, we can readily show that the degree of the polynomial \(Q_{N}^{[m]}(z)\)  is \(N(N+1)/2\) . Thus, Theorem 2 is proved.
